Unformatted text preview: v (t)|, the answer
is |v (2)| = 3m/sec.
2. x = 2t2 + 3 and y = t4 . Find dy
and 2 when t = −1.
dt = 4t3
|t=−1 = 1.
dx dx d dy
dt = 2t
2 d2 y
Therefore, 2 |t=−1 = .
3. Find the equation of the line normal to the curve x2 y 2 = 9 at the point (−1, 3).
By implicit diﬀerentiation and the Product rule,
x2 2 y
= −2xy 2
x 2y 2xy 2 + x2 2y dy
Therefore, the slope of the line tangent to the curve at (−1, 3) is
|(−1,3) = 3. We are,
however, looking for the equation of the line normal to the curve at this point, so the
1 2 1
slope of this line will be − , and the normal line equation is
y − 3 = − (x + 1).
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This note was uploaded on 02/10/2014 for the course MATH 235 taught by Professor Lucas during the Fall '08 term at James Madison University.
- Fall '08